# What is the Sum of the Polynomials

## Roots of a Polynomial

A “root” (or “zero”) is where the polynomial
is equal to zero:

Put but: a root is the x-value where the y-value equals zero.

## General Polynomial

If we have a full general polynomial similar this:

f(x) = axdue north
+ bxnorth-1
+ cxdue north-2
+ … + z

Then:

the roots gives
−b/a
• Multiplying
the roots gives:

• z/a
(for fifty-fifty degree polynomials similar quadratics)
• −z/a

(for odd degree polynomials like cubics)

Which can sometimes assistance us solve things.

How does this magic work? Let’s find out …

## Factors

We can take a polynomial, such as:

f(x) = axnorthward
+ bxn-1
+ cxn-2
+ … + z

And so factor it like this:

f(x) = a(x−p)(ten−q)(ten−r)…

And then p, q, r, etc are the
roots
(where the polynomial equals zero)

Let’due south endeavour this with a Quadratic (where the variable’s biggest exponent is 2):

ax2
+ bx + c

When the roots are
p
and

a(x−p)(10−q)

Is there a relationship between
a,b,c
and
p,q
?

Let’south expand
a(10−p)(x−q):

a(x−p)(x−q)

= a( ten2
− px − qx + pq )

= ax2
− a(p+q)ten + apq

Now let united states of america compare:

 Quadratic: ax2 +bx +c Expanded Factors: ax2 −a(p+q)x +apq

We tin now come across that
−a(p+q)x = bx, then:

−a(p+q) = b

p+q = −b/a

And
apq

= c
, so:

pq = c/a

And we get this result:

−b/a
• Multiplying the roots gives
c/a

This can aid u.s. answer questions.

### Example: What is an equation whose roots are five + √ii and 5 − √2

The sum of the roots is (5 + √2)  + (5 − √two) =
x

The product of the roots is (5 + √ii) (5 − √2) = 25 − two =
23

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And we desire an equation like:

ax2
+ bx + c = 0

When
a=1
we can work out that:

• Sum of the roots =

−b/a

=
-b
• Product of the roots =
c/a
=
c

Which gives us this result

tenii
− (sum of the roots)x + (production of the roots) = 0

The sum of the roots is 10, and product of the roots is 23, so we go:

xtwo
− 10x + 23 = 0

And here is its plot:

(Question: what happens if we cull
a=−1
?)

## Cubic

Now let the states look at a Cubic (i caste higher than Quadratic):

ax3
+ bx2
+ cx + d

As with the Quadratic, let us expand the factors:

a(ten−p)(x−q)(10−r)

= axthree
− a(p+q+r)xii
+ a(pq+pr+qr)x − a(pqr)

And we become:

 Cubic: ax3 +bx2 +cx +d Expanded Factors: axthree −a(p+q+r)xtwo +a(pq+pr+qr)x −apqr

We tin at present see that
−a(p+q+r)x2
= bxtwo
, so:

−a(p+q+r) = b

p+q+r = −b/a

And
−apqr

= d
, then:

pqr = −d/a

This is interesting … nosotros get the same sort of thing:

• Calculation the roots gives
−b/a

(exactly the aforementioned every bit the Quadratic)
• Multiplying the roots gives
−d/a
(like to +c/a for the Quadratic)

(We also become
pq+pr+qr = c/a, which can itself exist useful.)

## Higher Polynomials

The same pattern continues with higher polynomials.

In General: